Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

L1(f(s(s(y)), f(z, w))) → F(s(z), w)
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

L1(f(s(s(y)), f(z, w))) → F(s(z), w)
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.